The integral is : $\displaystyle \lim_{n\rightarrow \infty} \int_{\mathbb{R}^2}e^{-(x^2+y^2)^n}dxdy$

How could I solve this integral?

I've tried polar coordinates, but then I get $\displaystyle \lim_{n\rightarrow \infty} \int_{0}^{\infty}e^{-r^{2n}}r2\pi \,dr$, and I don't know how to solve this...

Any help would be appreciated.


It's $$2\pi \lim_{n\to\infty }\int_0^\infty re^{-r^{2n}}\mathrm d r$$ and not what you wrote. If you make the substitution $u=r^{2n}$, you'll get $$2\pi \lim_{n\to\infty }\int_0^\infty \frac{1}{2n}u^{\frac{1}{n}-1}e^{-u}du=\pi\lim_{n\to\infty }\frac{1}{n}\Gamma\left(\frac{1}{n}\right)=\pi\lim_{n\to\infty }\Gamma\left(\frac{1}{n}+1\right)\underset{(*)}{=}\pi\Gamma(1)=\pi.$$

$(*) :$ By continuity of $\Gamma$.

  • $\begingroup$ by the way, how would I solve this using the dominated convergence theorem? $\endgroup$ – An old man in the sea. Nov 14 '15 at 17:46
  • $\begingroup$ I made a correction. You can't use directly converge dominated here (if you could, your integral would make $0$). $\endgroup$ – Surb Nov 14 '15 at 17:56
  • $\begingroup$ That's what I thought, but the exercise asked to use the CDT... I guess there is a mistake in the exercise. Thanks ;) $\endgroup$ – An old man in the sea. Nov 14 '15 at 17:59
  • $\begingroup$ Sorry for the inconvenience, but if you put [0,infinity[ instead of R, then shouldn't you adjust also the domain for theta and r accordingly? $\endgroup$ – An old man in the sea. Nov 14 '15 at 18:15
  • 1
    $\begingroup$ You are totally right, and I apologize for this mistakes. You have to take $\theta\in [0,\pi/2[$ (but $r$ is still positif). My correction was useless, then I let the first version (i.e. with $\theta\in [0,2\pi[$). $\endgroup$ – Surb Nov 14 '15 at 18:35

Substituting $R = r^2$, we want the limit as $n\to\infty$ of

$$\pi \int_0^\infty e^{-R^n} dR$$

Note that

$$\int_0^1 e^{-R^n} dR \leq \int_0^\infty e^{-R^n} dR \leq \int_0^1 1 \ dR + \int_1^\infty e^{-R^n} dR$$

Now squeeze.


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